**1. Derive an expression for energy loss when two charged capacitors are connected with like plates together.**

When two capacitors charged at different potential differences and are connected by a metallic wire, there is flow of charge from the capacitor at higher potential to the capacitor at lower potential till the potential of the two capacitors becomes equal. In this process, the total charge remains constant but the total energy before joining the capacitors will be greater than that after joining. This means when two charged capacitors are joined by a wire, there is always a loss of energy.

Consider two charged capacitors of capacitances C_{1} and C_{2} that store the charges Q_{1} and Q_{2} respectively. Let V_{1} and V_{2} be the potentials developed on the respective capacitors.

_{1}before joining = ½ C1V

_{1}

^{2}

_{1}before joining = ½ C

_{2}V

_{2}

^{2}

_{1}= ½ C

_{1}V

_{1}

^{2}+ ½ C

_{2}V

_{2}

^{2 }….... (i)

_{1}+ Q

_{2}/ C

_{1}+ C

_{2}

_{1}V

_{1}+ C

_{2}V

_{2}/ C

_{1}+ C

_{2 }……. (ii)

_{2}= ½ (total capacity) * (common potential)

_{1}+ C

_{2}) * (C

_{1}V

_{1}+ C

_{2}V

_{2}/ C

_{1}+ C

_{2})

^{2}

_{1}V

_{1}+ C

_{2}V

_{2})

^{2}/ (C

_{1}+ C

_{2}) ……… (iii)

_{1}– E

_{2}= [½ C

_{1}V

_{1}

^{2}+ ½ C

_{2}V

_{2}

^{2}]– [ ½ (C

_{1}V

_{1}+ C

_{2}V

_{2})

^{2}/ (C

_{1}+ C

_{2})]

_{1}+ C

_{2})(C

_{1}V

_{1}

^{2}+ C

_{2}V

_{2}

^{2}) – (C

_{1}V

_{1}+ C

_{2}V

_{2})

^{2}] / 2(C

_{1}+ C

_{2})

_{1}C

_{2 }(V

_{1}

^{2}+ V

_{2}

^{2}– 2V

_{1}V

_{2 }) / 2(C

_{1}+ C

_{2})

_{1}– E

_{2}= C

_{1}C

_{2}(V

_{2}– V

_{1})

^{2 }/ 2(C

_{1}+ C

_{2})

This is the expression for the loss of energy after joining the charged capacitors.

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